Lf 23-43-63-2n3-kn2n-12- then k-

The correct option is C 2 2^3+4^3+6^3+…..+(2n)^3=2^3(1^3+2^3+3^3+….. #160;+n^3) =8n^2(n+1)^24 =2n^2(n+1)^2 Comparing it with =kn^2(n ...

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Byju's Answer

Standard XII

Mathematics

Permutation: n Different Things Taken r at a Time

lf 23+43+63...

Question

lf $2^{3} + 4^{3} + 6^{3} + \dots + (2 n)^{3} = k n^{2} (n + 1)^{2}$ , then $k =$

\frac{1}{2}

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\frac{3}{2}

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2

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Solution

The correct option is C $2$
$2^{3} + 4^{3} + 6^{3} + \dots . . + (2 n)^{3} = 2^{3} (1^{3} + 2^{3} + 3^{3} + \dots . . + n^{3})$
$= 8 \frac{n^{2} (n + 1)^{2}}{4}$
$= 2 n^{2} (n + 1)^{2}$
Comparing it with $= k n^{2} (n + 1)^{2}$ , we get
$k = 2$

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lf 23+43+63+…+(2n)3=kn2(n+1)2, then k=

The correct option is C 223+43+63+…..+(2n)3=23(13+23+33+…..+n3)=8n2(n+1)24=2n2(n+1)2Comparing it with =kn2(n+1)2, we getk=2

lf $2^{3} + 4^{3} + 6^{3} + \dots + (2 n)^{3} = k n^{2} (n + 1)^{2}$ , then $k =$

The correct option is C $2$
$2^{3} + 4^{3} + 6^{3} + \dots . . + (2 n)^{3} = 2^{3} (1^{3} + 2^{3} + 3^{3} + \dots . . + n^{3})$
$= 8 \frac{n^{2} (n + 1)^{2}}{4}$
$= 2 n^{2} (n + 1)^{2}$
Comparing it with $= k n^{2} (n + 1)^{2}$ , we get
$k = 2$